Names
Says something significant about a person; indicates what their destiny is
Lily vs Petunia
Remus Lupin
Sirius (dog star) Black (darkness)
Tom Marvolo Riddle -> I am Lord Voldemort; changing his destiny
Weasleys have Arthurian names
Draco (dragon, se
tk)[F(tk) F(tk1)] = + UF (f,P). (6) 35. *
Riemann-Stieltjes Integrals 327 Since (5) and
(6) hold for arbitrary yk in (tk1, tk), for each k,
we conclude4 UF (f,P) = n k=1 M(f,(tk1, tk)
[F(tk) F(tk1)] + U (fF, P). (7) and U (fF , P) =
n k=1 M(fF,(tk1, tk)(t
integrable, in which case the integrals are
equal. Proof The proof that F-integrability
implies Riemann-Stieltjes integrability imitates
the corresponding proof in Theorem 32.9. The
proof of the converse also imitates the
corresponding proof, but a little
very close to M(f,(tk1, tk), which appears in
the definition of UF (f,P) given after the
statement of Theorem 35.29. Similarly, we
concluded inequality (6) because we could
select each yk so that (fF)(yk) is very close to
M(fF,(tk1, tk), which appears in
|(s) (t)| = |s t| (1) provided there is no
even integer in the open interval (s, t). The
graph of is similar to that of g in Fig. 25.1 on
page 204; in fact, (x)=2g(x 2 ) for all x. The
continuous nowhere-differentiable function f
is defined by f(x) = n=1
to be 0. Assertion P1 holds by the hypothesis
of the lemma. Assume Pm holds for some m =
1,.,n 1. Since bm in Bm is arbitrary, Pm
implies Qm : am sup Bm K m 1 k=1 ak
sup Bk n k=m+1 akbk, for all choices of bk in
Bk, k = m + 1,.,n, but this is exactly asse
approach, we define E(x) = k=0 1 k! xk, (1)
and we define e = E(1). The series here has
radius of convergence + [Example 1, 23],
and E is differentiable on R [Theorem 26.5]. It
is easy [Exercise 26.5] to show E = E. The
fundamental property E(x + y) = E(x
F1(x) + limxt+ F2(x) = F1(t +) + F2(t +) with
similar identities for (F1 + F2)(t), (cF1)(t+) and
(cF1)(t). Hence for any partition P of [a, b],
we have UF1+F2 (f,P) = UF1 (f,P) + UF2 (f,P)
LF1+F2 (f,P) = LF1(f,P) + LF2 (f,P), (4) 35. *
Riemann-Stieltjes I
x>t, we have F(x) F(t) = cfw_cj : t 0, there is N so
that j=N+1 cj < . For x>t and x sufficiently
close to t, the sets cfw_uj : t< t1 < < tn = b such
that U(fF, P) L(fF , P) < 2 and UF (f,P) LF (f,P)
< 2 . (2) By the Mean Value Theorem 29.3 on
page 233 ap
suffices to show g f is F-integrable, because
then the sum (g f) + f would be F-integrable
by Theorem 35.8(ii). Thus it suffices to assume
g(x) = 0 except for finitely many points.
Consider any partition P = cfw_a = t0 < t1 < < tn
= b satisfying g(x)=0 fo
validity of (1) will not change if some of the uj
are not jump points so that cj = 0. So we may
assume a and b are in cfw_uj : 1 j m, and a =
u1 < u2 < < um = b. Let P be the partition cfw_a
= u1 < u2 < < um = b. Then we have F(u+ j )
F(u j ) = cj for j
about using the familiar exponential,
logarithmic and trigonometric functions in
examples and exercises. Most readers
probably found this an acceptable approach,
since they are comfortable with these basic
functions. In this section, we indicate three
way
subsequence of (xk) converges to some x in
[n, n]; to avoid subsequence notation, we will
assume the sequence (xk) itself converges to x.
Since xk x and fk f uniformly, we have
fk(xk +h) f(x+h) for each h [0, 1 n); see
Exercise 24.17 on page 200. For each
0, this proves (2) so that f is not differentiable
at a. 350 7. Capstone 38.2 Remark. For later
use, note the function in the preceding
proof satisfies supcfw_|(y)| : y R = 2, and for
any y in R we have |(y + h) (y)| = h for
sufficiently small h > 0. (1)
yz = E(L(y)E(L(z) = E(u)E(v). Assertion (iii)
follows from (iv) of Theorem 37.4 [Exercise
37.2]. Consider b > 0 and r Q, say r = m n
where m, n Z and n > 0. It is customary to
write br for that positive number a such that
an = bm. By (ii) of Theorem 37.4,
Show limxt F(x) exists for t in (a, b] and is
equal to supcfw_F(x) : x (a, t). (b) Show
limxt+ F(x) exists for t in [a, b) and is equal
to infcfw_F(x) : x (t, b). 35.2 Calculate " 3 0 x2
dF(x) for the function F in Example 4. 35.3 Let F
be the step functi
bounded on [a, b]. For > 0 we apply Lemma
35.15 to obtain P = cfw_a = t0 < t1 < < tn = b
where F(t k ) F(t + k1) < f(b) f(a) for k = 1,
2,.,n. Since M(f,(tk1, tk) = f(t k ) f(tk) and
m(f,(tk1, tk) = f(t + k1) f(tk1), we have UF
(f,P) LF (f,P) n k=1 [f(tk)
integrating the function f(n) n times. At each
step, one obtains a polynomial of one higher
degree; see Corollary 29.5 and Exercise 29.7.
We write C(, ) for the space of functions
f infinitely differentiable on (, ), i.e., all the
derivatives f(n) exist o
Now by Lemma 35.30, we have |UF (f,P) U
(fF, P)| , so by the triangle inequality, we
have |UF (f) U(fF )| < 3. Since is arbitrary, UF
(f) = U(fF). Likewise LF (f) = L(fF). 4Some
readers may appreciate more explanation. To
avoid breaking the flow of the pr
Integrals 323 35.27 Proposition. If f is bounded
on [a, b], then UF (f) = UF (f) and LF (f) = LF (f).
Thus f is integrable using closed intervals in
the Darboux-Stieltjes definitions if and only if
it is integrable using open intervals, and in this
case t
=1= F(1)F(1) F(1)F(1). The generalization
does hold provided the functions in the
integrands take the middle values at each of
their jumps, as we next prove. The result is a
special case of a theorem given by Edwin
Hewitt [29]. 35.19 Theorem [Integration
complex values since sin x = 1 2i [eix eix],
etc.; see [62, Chap. 8]. A development of the
trigonometric functions analogous to approach
37.3 can proceed as follows. Since we believe
arcsin x = ! x 0 (1 t 2) 1/2 dt, we can define
A(x) as this integral and
the slope exceeds 2n. Let m = a+b 2 be the
midpoint of the interval [a, b]. Then we have
(b) (m) b a = (m) (a) b a > n; thus there
exists > 0 so that (b) [ (m) + ] b a > n and
(m) [ (a) + ] b a > n. (1) The desired
rhomboid R is bounded above and below by
show s< uj t, which we write as cfw_t0 < t1 <
< tn where s 0, then there exists a partition P
= cfw_a = t0 < t1 < < tn = b 314 6. Integration
such that F(t k ) F(t + k1) < for k = 1, 2, . . . ,
n. (1) Proof The conclusion (1) does not
depend on the exact
35.2 on page 300. The usual Riemann-Stieltjes
integral is defined via the sums SF (f,P) = n
k=1 f(xk)[F(tk) F(tk1)], where xk is in [tk1,
tk], and the mesh is defined in Definition 32.6;
compare Definition 35.24. The usual RiemannStieltjes integrability c
we conclude L(xr) = rL(x) for r Q and x > 0.
(3) For b > 0 and rational r, (3) implies br =
E(L(br ) = E(rL(b). 342 7. Capstone Because of
this, we define bx = E(xL(b) for x R. The
familiar properties of exponentials and their
inverses [logarithms!] are n
U(f,P) for upper and lower Darboux sums could
have used open intervals (tk1, tk) instead of
closed intervals [tk1, tk]. The resulting 35. *
Riemann-Stieltjes Integrals 325 Darboux sums
L (f,P) and U (f,P) might be different. But if we
define U (f) = infcf
decreasing. (iii) If b = 1, then B maps R onto
(0,). (iv) bu+v = bubv for u, v R. Proof
Exercise 37.3. If b = 1, the function B has an
inverse function. 37.9 Definition. For b > 0 and
b = 1, the inverse of B(x) = bx is written logb.
The domain of logb is
and " d c g dG = " b a f dF. 35.12 Let (uj ) be an
enumeration of the rationals in [a, b] and let
(cj ) be a sequence of positive integers such
that cj < . (a) Show F = cjJuj defines a strictly
increasing function on [a, b]. (b) At what
points is F contin
35.7 Let f and g be F-integrable functions on [a,
b]. Show (a) f 2 is F-integrable. (b) fg is Fintegrable. (c) max(f, g) and min(f, g) are Fintegrable. 35.8 Let g be continuous on [a, b]
where g(x) 0 for all x [a, b] and define F(t)
= " t a g(x) dx for t